The runner has initial velocity vector
![\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v_0%3D%5Cleft%282.88%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29%5C%2C%5Cvec%5Cjmath)
and acceleration vector
![\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)](https://tex.z-dn.net/?f=%5Cvec%20a%3D%5Cleft%280.350%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%28%5Ccos%28-52.0%5E%5Ccirc%29%5C%2C%5Cvec%5Cimath%2B%5Csin%28-52.0%5E%5Ccirc%29%5C%2C%5Cvec%5Cjmath%29)
so that her velocity at time
is
![\vec v=\vec v_0+\vec at](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cvec%20v_0%2B%5Cvec%20at)
She runs directly east when the vertical component of
is 0:
![2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s](https://tex.z-dn.net/?f=2.88%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%2B%5Cleft%280.350%5C%2C%5Cdfrac%7B%5Crm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5Csin%28-52.0%5E%5Ccirc%29%5C%2Ct%3D0%5Cimplies%20t%3D10.4%5C%2C%5Crm%20s)
It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time
would be
![\vec x=\vec v_0t+\dfrac12\vec at^2](https://tex.z-dn.net/?f=%5Cvec%20x%3D%5Cvec%20v_0t%2B%5Cdfrac12%5Cvec%20at%5E2)
so that after 10.4 s, her position would be
![\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20x%3D%2810.1%5C%2C%5Cmathrm%20m%29%5C%2C%5Cvec%5Cimath%2B%2817.2%5C%2C%5Cmathrm%20m%29%5C%2C%5Cvec%5Cjmath)
which is 19.9 m away from her starting position.
Answer:
<em>The cyclist is traveling at 130 m/s</em>
Explanation:
<u>Constant Acceleration Motion
</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
![v_f=v_o+at](https://tex.z-dn.net/?f=v_f%3Dv_o%2Bat)
The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.
Apply the above equation:
![v_f=10+6\cdot 20](https://tex.z-dn.net/?f=v_f%3D10%2B6%5Ccdot%2020)
![v_f=10+120](https://tex.z-dn.net/?f=v_f%3D10%2B120)
![v_f=130\ m/s](https://tex.z-dn.net/?f=v_f%3D130%5C%20m%2Fs)
The cyclist is traveling at 130 m/s
Total distance = 76+54 = 130km
total time = 2+5 = 7hrs
Av. speed = 130/7 = 18.571km/hr = 18.6 km/h ( 3 sig fig)
Quantum entanglement<span> is a physical phenomenon that occurs when pairs or groups of particles are generated or interact in ways such that the </span>quantum<span> state of each particle cannot be described independently — instead, a </span>quantum<span> state must be described for the system as a whole.</span>